Question

How do you find a vector that is orthogonal to 5i + 12j ?

Answer 1

`\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad \cfrac{a}{b}\\\\ slope=\cfrac{a}{{{ b}}}\qquad negative\implies -\cfrac{a}{{{ b}}}\qquad reciprocal\implies - \cfrac{{{ b}}}{a}\\\\ -------------------------------\\\\`


`\bf \boxed{5i+12j}\implies \begin{array}{rllll} \ \textless \ 5&,&12\ \textgreater \ \\ x&&y \end{array}\quad slope=\cfrac{y}{x}\implies \cfrac{12}{5} \\\\\\ slope=\cfrac{12}{{{ 5}}}\qquad negative\implies -\cfrac{12}{{{ 5}}}\qquad reciprocal\implies - \cfrac{{{ 5}}}{12} \\\\\\ \ \textless \ 12, -5\ \textgreater \ \ or\ \ \textless \ -12,5\ \textgreater \ \implies \boxed{12i-5j\ or\ -12i+5j}`

if we were to place <5, 12> in standard position, so it'd be originating from 0,0, then the rise is 12 and the run is 5.

so any other vector that has a negative reciprocal slope to it, will then be perpendicular or "orthogonal" to it.

so... for example a parallel to <-12, 5> is say hmmm < -144, 60>, if you simplify that fraction, you'd end up with <-12, 5>, since all we did was multiply both coordinates by 12.

or using a unit vector for those above, then


`\bf \textit{unit vector}\qquad \cfrac{\ \textless \ a,b\ \textgreater \ }\implies \cfrac{\ \textless \ a,b\ \textgreater \ }{√(a^2+b^2)}\implies \cfrac{a}{√(a^2+b^2)},\cfrac{b}{√(a^2+b^2)} \\\\\\ \cfrac{12,-5}{√(12^2+5^2)}\implies \cfrac{12,-5}{13}\implies \boxed{\cfrac{12}{13}\ ,\ \cfrac{-5}{13}} \\\\\\ \cfrac{-12,5}{√(12^2+5^2)}\implies \cfrac{-12,5}{13}\implies \boxed{\cfrac{-12}{13}\ ,\ \cfrac{5}{13}}`